The interval in which the function `f(x) = 2 x^(3)+ 3x^(2) - 12 x + 1 ` is xtrictly increasing is

Options

A. [-2,1]
B. `(- infty, - 2] cup [1, infty)`
C. `(- infty, 1]`
D. `(- infty, - 1] cup [2, infty)`

Answer

To determine the intervals in which the function \( f(x) = 2x^3 + 3x^2 - 12x + 1 \) is strictly increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \) with respect to \( x \).

\[ f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 12x + 1) \]

Using the power rule of differentiation, we get:

\[ f'(x) = 6x^2 + 6x - 12 \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:

\[ 6x^2 + 6x - 12 = 0 \]

Dividing the entire equation by 6 simplifies it:

\[ x^2 + x - 2 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x + 2)(x - 1) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor equal to zero gives us the critical points:

\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 5: Determine the intervals
The critical points divide the number line into intervals. We will test the sign of \( f'(x) \) in each interval:

1. \( (-\infty, -2) \)
2. \( (-2, 1) \)
3. \( (1, \infty) \)

### Step 6: Test the intervals
We can choose test points from each interval to determine if \( f'(x) \) is positive or negative.

- **Interval \( (-\infty, -2) \)**: Let’s choose \( x = -3 \)
\[ f'(-3) = 6(-3)^2 + 6(-3) - 12 = 54 - 18 - 12 = 24 \quad (\text{positive}) \]

- **Interval \( (-2, 1) \)**: Let’s choose \( x = 0 \)
\[ f'(0) = 6(0)^2 + 6(0) - 12 = -12 \quad (\text{negative}) \]

- **Interval \( (1, \infty) \)**: Let’s choose \( x = 2 \)
\[ f'(2) = 6(2)^2 + 6(2) - 12 = 24 + 12 - 12 = 24 \quad (\text{positive}) \]

### Step 7: Conclusion
From our tests, we find:
- \( f'(x) > 0 \) in the intervals \( (-\infty, -2) \) and \( (1, \infty) \)
- \( f'(x) < 0 \) in the interval \( (-2, 1) \)

Thus, the function \( f(x) \) is strictly increasing in the intervals:

\[ (-\infty, -2) \cup (1, \infty) \]

### Final Answer
The function \( f(x) \) is strictly increasing in the intervals \( (-\infty, -2) \) and \( (1, \infty) \).

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